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3 years ago

A bar magnet is held in place while another bar magnet is placed near it. The second bar magnet spins around and

Physics

1 answer:

Vinil7 [7]3 years ago

4 0

Answer:

Explanation:

the increase energy stored in thw system is proportional to the decrease in kinetic energy

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What is the magnitude (in N/C) and direction of an electric field that exerts a 4.00 x 10^−5 N upward force on a −2.00 µC charge

emmasim [6.3K]

Answer:

The magnitude of electric field is $1.25\times10^{14}\ N/C$ in downward.

Explanation:

Given that,

Force $F= 4.00\times10^{-5}\ N$

Charge q= -2.00 μC

We know that,

Charge is negative, then the electric field in the opposite direction of the exerted force.

We need to calculate the magnitude of electric field

Using formula of electric force

$F = qE$

$E = \dfrac{F}{q}$

$E=\dfrac{4.00\times10^{-5}}{-2.00\times1.6\times10^{-19}}$

$E=-1.25\times10^{14}\ N/C$

Negative sign shows the opposite direction of electric force.

Hence, The magnitude of electric field is $1.25\times10^{14}\ N/C$ in downward.

8 0

4 years ago

Why do faults often occur along plate boundaries?

Slav-nsk [51]

Earthquakes can also occur far from the edges of plates, along faults. Faults are cracks in the earth where sections of a plate (or two plates) are moving in different directions. Faults are caused by all that bumping and sliding the plates do. They are more common near the edges of the plates.

4 0

3 years ago

Un ciclista recorre 10 km en 2 h.Calcula su velocidad media en km/h.¿cuantos metros recorre cada segundo

lys-0071 [83]

Answer:

La velocidad media es 5 $\frac{km}{h}$ , que equivale a 1.389 $\frac{m}{s}$

Explanation:

La velocidad es una magnitud física que expresa la relación entre el espacio recorrido por un objeto y el tiempo empleado para ello.

La velocidad media relaciona el cambio de la posición con el tiempo empleado en efectuar dicho cambio. Por lo que se calcula como la distancia recorrida por un objeto dividido por el tiempo transcurrido:

$velocidad=\frac{distancia}{tiempo}$

En este caso:

distancia= 10 km= 10,000 m (siendo 1 km= 1,000 m)
tiempo= 2 h= 7,200 s (siendo 1 h= 3,600 s)

Entonces, reemplazando en la definición de velocidad media:

$velocidad=\frac{10 km}{2 h}= \frac{10,000 m}{7,200 s}$

Resolviendo se obtiene:

$velocidad=5 \frac{km}{h}= 1.389\frac{m}{s}$

La velocidad media es 5 $\frac{km}{h}$ , que equivale a 1.389 $\frac{m}{s}$

4 0

3 years ago

Calculate the electric field at the center of a square

pantera1 [17]

Answer:

$E_y=1175510.2\ N.C^{-1}$

The Magnitude of electric field is in the upward direction as shown directly towards the charge $q_1$ .

Explanation:

Given:

side of a square, $a=52.5\ cm$

charge on one corner of the square, $q_1=+45\times 10^{-6}\ C$

charge on the remaining 3 corners of the square, $q_2=q_3=q_4=-27\times 10^{-6}\ C$

Distance of the center from each corners $=\frac{1}{2} \times diagonals$

$diagonal=\sqrt{52.5^2+52.5^2}$

$diagonal=74.25\cm=0.7425\ m$

∴Distance of center from corners, $b=0.3712\ m$

Now, electric field due to charges is given as:

$E=\frac{1}{4\pi\epsilon_0}\times \frac{q}{b^2}$

For charge $q_1$ we have the field lines emerging out of the charge since it is positively charged:

$E_1=9\times 10^9\times \frac{45\times 10^{-6}}{0.3712^2}$

$E_1=2938775.5\ N.C^{-1}$

Force by each of the charges at the remaining corners:

$E_2=E_3=E_4=9\times 10^9\times \frac{27\times 10^{-6}}{0.3712^2}$

$E_2=E_3=E_4=1763265.3\ N.C^{-1}$

Now, net electric field in the vertical direction:

$E_y=E_1-E_4$

$E_y=1175510.2\ N.C^{-1}$

Now, net electric field in the horizontal direction:

$E_y=E_2-E_3$

$E_y=0\ N.C^{-1}$

So the Magnitude of electric field is in the upward direction as shown directly towards the charge $q_1$ .

8 0

3 years ago

A tennis player hits a ball at an angle of 50 degrees above horizontal so that it has an acceleration of 12 m/s2. What is the ho

harkovskaia [24]

So we want to know what is the magnitude of the horizontal component of acceleration ah if we know that the overall acceleration a=12 m/s^2 and the angle of overall acceleration and the horizontal acceleration is α=50°. We know that ah=a*cosα. So now it isn't hard to get the horizontal component: ah=12*cos50=12*0.64=7.71 m/s^2. So the correct answer is ah=7.71 m/s^2.

6 0

3 years ago